Write a real world problem that involves classifying a quadrilateral then solve the problem.

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An amusement park ride has a moving platform attached to four swinging arms. The platform swings back and forth, higher and higher, until it goes over the top and around in a circular motion. In the diagram below, AD and BC represent two of the swinging arms, and DC <span>is parallel to the ground
</span>(line l). Explain why the moving platform AB <span>is always parallel to the ground.</span>

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PLEASE HELP, BEEN STUCK FOR AN HOUR

djverab [1.8K]

Answer:

Step-by-step explanation:

To sketch a quadratic function we need two things:

1) Nature of the curve

2) Vertex

3) y-intercept

The completing square form of the quadratic equation is:

$y=a(x-h)^2+k$

where,

a represents the nature of the graph it can be maximum or minimum , meaning, if a > 0 then minimum(u shaped curve/happy face) and if a < 0 then maximum(n shaped curve/sad face).

h represents the x-coordinate of the vertex.

k represents the y-coordinate of the vertex.

Now if we compare g(x) with our completing square form we get the following:

$g(x)=(x-4)^2+12\\y=a(x-h)^2+k\\$

When we simply compare the following we get ,

a = 1 , which means a > 0 since 1 is greater than 0 the nature of the curve will be minimum(happy face/u shaped)

h = 4, which means the x-coordinate of the vertex is 4

k = 12, which means the y-coordinate of the vertex is 12

Now we have the nature of the curve, we have the vertex now all we need is the y-intercept.

For y-intercept:

For y-intercept meaning at which point will the graph cross the y-axis(0 , y)

For that we expand the formula and turn it into the standard quadratic equation form by using the formula (a - b)^2

$y=(x-4)^2+12\\y=((x)^2-2(x)(4)+(4)^2)+12\\y=x^2-8x+16+12\\y=x^2-8x+28\\$

now we compare with the standard quadratic form:

$y=ax^2+bx+c$

here c is the y-intercept and while comparing we can see that c = 28 ,

so the curve cuts the y-axis at (0 , 28)

So we have all the three things that we need to graph our function.

So we just plot the y-intercept , the vertex , and join the dots. Just a tip draw a dotted line on the x-coordinate of the vertex because the vertex point is also called as a turning point where the graph goes in the opposite direction just like a mirror reflection. I attached 2 images you can check them out. One is handmade(i know i suck at drawing but still xD) , one is sketched by online graphing calculator.